 Solution

Divide the equation by $\sqrt{1-x^2}$ to obtain its Sturm-Liouville form:Thus $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. For $L=-w^{-1}(pd/dx{)}^{\prime }$, the Lagrange identity for a Sturm-Liouville operator gives the endpoint form $[p({\bar{f}}^{\prime }g-\bar{f}{g}^{\prime }){]}_{-1}^1$. A natural self-adjoint choice isUnder $x=\cos \theta$, these become the ordinary zero-derivative conditions at $\theta =0,\pi$ for $-d^2/d{\theta }^2$. They therefore define the realization whose Neumann eigenfunctions of the Chebyshev differential operator are $T_n(x)=\cos (n\arccos x)$, with $n=0,1,\dots$.

Boundedness alone does not choose these endpoint conditions: $\sin (n\arccos x)$ is also bounded but generally has a nonzero endpoint value of $p{y}^{\prime }$. Requiring bounded $y^{\prime }$ is a sufficient stronger regularity condition for the displayed zero-flux limits. Other self-adjoint endpoint choices are possible; the boundary form and the operator domain must specify which is intended.